9.5 Inductance of a Solenoid
Let’s assume that we have a solenoid, something like this, with N number of turns, and we let current i flow through this solenoid, and as a result of that we generate a magnetic field B, filling the region surrounded by the solenoid. We’d like to calculate the inductance of the solenoid, and therefore we will follow the procedure that we have introduced earlier.
First, we assume that the certain amount of current is flowing through the solenoid. As the second step, we will calculate the magnetic field of this solenoid. We will assume that the solenoid that we’re dealing with is a perfect solenoid, and let’s assume that its total length is given as l, length of solenoid, and let’s assume that little n, which is equal to number of turns per unit length, gives us the number density of the solenoid, number density of turns.
All right. In order to calculate the magnetic field, we’re going to apply Ampere’s Law. As a matter of fact, we did this earlier step by step, but let’s do it one more time as a review. If we just cut the solenoid vertically down, we end up with an upper branch and a lower branch, and if the upper branch is carrying the current out of plane direction, then the lower branch will carry the current into the plane direction. In other words, if the current is coming out of here, then it is going to go into the plane here and then it will come out here and go into the plane there, and so on and so forth.
In such a current configuration as we have seen earlier, it’s going to generate a magnetic field along the axis of the solenoid, and fill the interior region of the solenoid. Now, for a perfect solenoid, as you recall, this is the case that l is much, much greater than the, let’s say, diameter of the solenoid. Then one can assume that Bout, the magnetic field outside of the solenoid, is 0. Of course, this is not the actual case, but comparing the strength of the solenoid inside relative to the one outside, one can actually make this approximation.
Let’s assume that we’re interested with the magnetic field at this point P, an arbitrary point inside of the solenoid, and to be able to calculate the magnetic field, first we look at the field line passing through that point. So it’s going to be something like this. And we’re going to choose a closed loop such that one side of this loop will coincide with the magnetic field line passing through the point of interest.
So, we’ll choose a rectangular loop in this case, of this shape, and if we just say this side is the segment one and this one is segment two, segment three, and segment four, the Ampere’s Law, which says integral of B dot dl over a closed contour C is equal to μ0 times i-enclosed. This closed loop integral can be expressed as the open path integrals of the four sides of this rectangular loop, and then they are added together to be able to get the whole closed loop integral.
In other words, we can express this as integral over the first segment of B dot dl plus integral over the second segment of B dot dl plus integral over the third segment of B dot dl and plus integral over the fourth segment of B dot dl. They all added to closed path integral or closed loop integral and will be equal to μ0 times i-enclosed.
Now, if we look at each one of these segments separately, for the first segment, dl is pointing down and magnetic field is pointing to the right. Therefore the angle between them will be 90 degrees inside of the solenoid. If we look at the outside of the solenoid, here Bout is 0, therefore from the outside part of the first segment, since magnetic field is 0, there will not be any contribution, and for the inside region of the first segment, the angle between B and dl is 90 degrees, so this is going to give us 0. And reason, that the